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Limit of a function | IIT JAM 2017 | Problem 8

Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8


Let

    \[f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0\]


Write L=\displaystyle\lim_{x \to 0^{-}} f(x) and R=\displaystyle\lim_{x \to 0^{+}} f(x) .

Then which one of the following is true?

  • L exists but R does not exist
  • L does not exist but R exists
  • Both L and R exist
  • Neither L nor R exists

Key Concepts


Real Analysis

Function

Limit

Check the Answer


Answer: L exists but R does not exists

IIT JAM 2017 , Problem 8

Try with Hints


Given that, f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0

therefore,

f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0

f(x)=\bigg\{\begin{array}{cc} (2+x) \sin \left(\frac{1}{x}\right),  & , x>0 \\ -x \sin \left(\frac{1}{x}\right),  & x<0  \\ \end{array}

Let, L=\displaystyle\lim_{x \to 0^{-}} f(x)

and , R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .

Now,

L= \displaystyle\lim_{x \to 0^{-}} f(x)

\quad = \displaystyle\lim_{x \to 0^{-}}   -x \sin \left(\frac{1}{x}\right)

\quad = -\displaystyle\lim_{x \to 0^{-}}   x \sin \left(\frac{1}{x}\right)

Theorem : If D \subset \mathbb R and f,g : D \to \mathbb R . Let c \in D. If f is bounded on N'(c)\cup D and \displaystyle\lim_{x \to c} g(x)=0, then \displaystyle\lim_{x \to c}(f.g)(x)=0.

Now , \sin \left(\frac{1}{x}\right) is bounded in \mathbb R - \{0\} and \displaystyle\lim_{x \to 0^{-}} x=0 , then \displaystyle\lim_{x \to 0^{-}} f(x) exists and equal to 0.

But,

R=\displaystyle\lim_{x\to 0^{+}}f(x)

\quad =  \displaystyle\lim_{x\to 0^{+}}  (2+x) \sin \left(\frac{1}{x}\right),

\quad=  \displaystyle\lim_{x\to 0^{+}} 2\sin  \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right)

\lim_{x \to 0^{+}} \sin  \left(\frac{1}{x}\right) does not exists [Why?]

Then L exists but R does not.

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Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8


Let

    \[f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0\]


Write L=\displaystyle\lim_{x \to 0^{-}} f(x) and R=\displaystyle\lim_{x \to 0^{+}} f(x) .

Then which one of the following is true?

  • L exists but R does not exist
  • L does not exist but R exists
  • Both L and R exist
  • Neither L nor R exists

Key Concepts


Real Analysis

Function

Limit

Check the Answer


Answer: L exists but R does not exists

IIT JAM 2017 , Problem 8

Try with Hints


Given that, f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0

therefore,

f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0

f(x)=\bigg\{\begin{array}{cc} (2+x) \sin \left(\frac{1}{x}\right),  & , x>0 \\ -x \sin \left(\frac{1}{x}\right),  & x<0  \\ \end{array}

Let, L=\displaystyle\lim_{x \to 0^{-}} f(x)

and , R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .

Now,

L= \displaystyle\lim_{x \to 0^{-}} f(x)

\quad = \displaystyle\lim_{x \to 0^{-}}   -x \sin \left(\frac{1}{x}\right)

\quad = -\displaystyle\lim_{x \to 0^{-}}   x \sin \left(\frac{1}{x}\right)

Theorem : If D \subset \mathbb R and f,g : D \to \mathbb R . Let c \in D. If f is bounded on N'(c)\cup D and \displaystyle\lim_{x \to c} g(x)=0, then \displaystyle\lim_{x \to c}(f.g)(x)=0.

Now , \sin \left(\frac{1}{x}\right) is bounded in \mathbb R - \{0\} and \displaystyle\lim_{x \to 0^{-}} x=0 , then \displaystyle\lim_{x \to 0^{-}} f(x) exists and equal to 0.

But,

R=\displaystyle\lim_{x\to 0^{+}}f(x)

\quad =  \displaystyle\lim_{x\to 0^{+}}  (2+x) \sin \left(\frac{1}{x}\right),

\quad=  \displaystyle\lim_{x\to 0^{+}} 2\sin  \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right)

\lim_{x \to 0^{+}} \sin  \left(\frac{1}{x}\right) does not exists [Why?]

Then L exists but R does not.

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